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Question

A \( 55 \, \text{mH} \) inductor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) source.
What is the peak current?

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Explanation

**Inductive reactance** X_L = ω L =2π f L (Ω), L inductance (H), f frequency (Hz), ω=2πf angular frequency (rad/s), opposes AC, impedance Z = X_L for pure L, current lags voltage by 90°, I_rms = V_rms/X_L, I_peak = V_peak/X_L. For 85 mH, 50 Hz, X_L=2π×50×0.085=26.7 Ω, V_rms=230 V, I_rms=8.61 A, I_peak=12.18 A. X_L = ω L , ω = 2π × 60 = 376.8 rad/s . L = 55 × 10⁻³ H . X_L = 376.8 × 0.055 = 20.72 Ω . RMS current: I = (V/X_L) = (110/20.72) ≈ 5.31 A . Peak current: i_m = √(2) I = 1.414 × 5.31 ≈

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